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stop & restart clock

Charanraj Mohan

Hi,

Just a question. Can b silly. How to stop the clock and start again in a transient time ?? Say, the whole clock time has same period, pulse width, tr, tf, td. I can do it in Vpwl, but the clock I am planning has more cycles :(

Thanks in advance.

Parents

  • One way would be to use a bit source (so use source from analogLib and set the type to be "bit"). You can then either define a sequence of 1 and 0 values, along with the values these correspond to, rise times, periods and so on, or you can reference a "pattern" which could be defined in an include file. See "spectre -h pattern" and "spectre -h vsource". The pattern allows nested patterns with repetitions - so you could define a fairly complex sequence simply:

    // example of using pattern to simply define a clock pulse that
    // is enabled for 50 periods, then off for 20 periods, then on for another
    // 17 periods

    parameters clkFreq=1M

    clk50 pattern data="10" rpttimes=49 // means 50 total
    off20 pattern data="00" rpttimes=19 // means 20 total
    clk17 pattern data="10" rpttimes=16 // means 17 total

    vclk (clk 0) vsource type=bit data="00,clk50,off20,clk17,0000" val0=0 \
    val1=1.8 rise=.1n fall=0.1n period=1/(clkFreq*2)

    tran tran stop=(50+20+17+3)*1/clkFreq

    Here's what it ends up looking like:

    Hope that helps!

    Regards,

    Andrew.

Reply

  • One way would be to use a bit source (so use source from analogLib and set the type to be "bit"). You can then either define a sequence of 1 and 0 values, along with the values these correspond to, rise times, periods and so on, or you can reference a "pattern" which could be defined in an include file. See "spectre -h pattern" and "spectre -h vsource". The pattern allows nested patterns with repetitions - so you could define a fairly complex sequence simply:

    // example of using pattern to simply define a clock pulse that
    // is enabled for 50 periods, then off for 20 periods, then on for another
    // 17 periods

    parameters clkFreq=1M

    clk50 pattern data="10" rpttimes=49 // means 50 total
    off20 pattern data="00" rpttimes=19 // means 20 total
    clk17 pattern data="10" rpttimes=16 // means 17 total

    vclk (clk 0) vsource type=bit data="00,clk50,off20,clk17,0000" val0=0 \
    val1=1.8 rise=.1n fall=0.1n period=1/(clkFreq*2)

    tran tran stop=(50+20+17+3)*1/clkFreq

    Here's what it ends up looking like:

    Hope that helps!

    Regards,

    Andrew.

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